TheSmartOne
  • TheSmartOne
Evaluate the limit as x approaches 0 of (1 - x^(sin(x)))/(x*log(x)) aka: \(\frac{1-x^{sin(x)}}{x*log(x)}\) Please include steps/explanation.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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TheSmartOne
  • TheSmartOne
@ganeshie8 @Michele_Laino
Er.Mohd.AMIR
  • Er.Mohd.AMIR
use l'Hospital rrule
TheSmartOne
  • TheSmartOne
\(\LARGE \frac{1-x^{sin(x)}}{x*log(x)}\)

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More answers

TheSmartOne
  • TheSmartOne
\(\large \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } =_{LH} \lim_{x \to 0^{+}} \frac{0 -x^x( 1 + \log (x)) }{1 + \log (x) } \\ = \large \lim_{x \to 0^{+}} (-x^x) = \large - \lim_{x \to 0^{+}} (x^x) = -1\)
TheSmartOne
  • TheSmartOne
@Er.Mohd.AMIR correct?
TheSmartOne
  • TheSmartOne
@sleepyhead314 I know you know the answer ;)
jango_IN_DTOWN
  • jango_IN_DTOWN
OK I can solve the problem
sleepyhead314
  • sleepyhead314
cannot directly use l'hopital
TheSmartOne
  • TheSmartOne
yea, before that I had this; \(\large \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{x \log x } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{\sin x} }{ \log( x^x) } \\~\\ \large = \lim_{x \to 0^{+}} \frac{1- x^{x} }{ \log( x^x) } ~~ \normalsize{\text{ substituting x for sin x } } \\~\\ \large = \frac{\lim_{x \to 0^{+}} (1) - \lim_{x \to 0^{+}} \left( x^{x}\right) }{ \log( \lim_{x \to 0^{+}}x^x) } = \frac{1-1}{\log(1)} = \frac{0}{0}\)
sleepyhead314
  • sleepyhead314
I know that you know that you have absolutely no idea what that means
TheSmartOne
  • TheSmartOne
or do I (;
jango_IN_DTOWN
  • jango_IN_DTOWN
1 Attachment
Er.Mohd.AMIR
  • Er.Mohd.AMIR
wrong diff of numerator.
jango_IN_DTOWN
  • jango_IN_DTOWN
check my solution
TheSmartOne
  • TheSmartOne
@hartnn :)
anonymous
  • anonymous
calculus problem lol
just_one_last_goodbye
  • just_one_last_goodbye
@TheSmartOne did you tag Hartnn for help or for CoC enforcement? Cause I can help with the question.
TheSmartOne
  • TheSmartOne
for help :)
jango_IN_DTOWN
  • jango_IN_DTOWN
@ganeshie8 does my solution contains error? do let me know
TheSmartOne
  • TheSmartOne
if you can help @just_one_last_goodbye , go for it :)
just_one_last_goodbye
  • just_one_last_goodbye
working it out on my paper :)
TheSmartOne
  • TheSmartOne
sure, tell me what you get :)

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